My proof on sum of two sequences is real number? Let $\lim_{n\to\infty}a_n$ and $\lim_{n\to\infty}b_n$ be real numbers. Show that their sum
$\lim_{n\to\infty}(a_n+b_n)$ is also a real number. Hint: you can use the triangle inequality.
This is what I came up with so far:
So lets say $\lim_{n\to\infty}a_n=x$. Then for every $\varepsilon>0$, there exists an $N$ such that $|a_n-x|<\varepsilon$ for every $n\geq N$. Then if $\lim_{n\to\infty}b_n=y$ (where $x$ and $y$ are real numbers, given in question) for every $\varepsilon>0$ we can find an $N$ such that $|b_n-y|<\varepsilon$ for every $n\geq N$.
We have $|a_n-x +b_n-y|<\varepsilon$ where $\varepsilon>0$ and $n\geq N$.
Now with the triangle inequality $|x+y|\leq |x| + |y|$ that implies
$|a_n-x +b_n-y|\leq |a_n-x| + |b_n-y|$ so if we can make
$|a_n-x| + |b_n-y| <\varepsilon$ we know that $|a_n-x +b_n-y|$ is also smaller then $\varepsilon$.
but I dont know how to conclude the proof, that this sum is also a real number. I can make both terms on the right $\varepsilon/2$, but is this proof that their sum is also a real number?
 A: You're very much on the right track, but some corrections are needed. For example, you can say that

We have $|a_n−x+b_n−y|<\varepsilon$ where $\varepsilon>0$ and $n\geq N$.

No, we don't have that — this is what we need to prove, not what we know.

So lets say $\lim\limits_{n\to\infty}a_n=x$. Then for every $\varepsilon>0$, there exists an $N$ such that $|a_n−x|<\varepsilon$ for every $n\geq N$. Then if $\lim\limits_{n\to\infty}b_n=y$ (…) for every $\varepsilon>0$ we can find an $N$ such that $|b_n−y|<\varepsilon$ for every $n\geq N$.

While both statements are correct if stated separately, once you put them together, it doesn't look right. Using the same "$N$" gives an impression that it is the same number, while it doesn't have to be, i.e. not immediately from the definition. So you have to take care of this little issue, for example like that:
Assume $\lim\limits_{n\to\infty}a_n=x$. Then for every $\varepsilon>0$, there exists an $N_a$ such that $|a_n−x|<\varepsilon$ for every $n\geq N_a$. Assume $\lim\limits_{n\to\infty}b_n=y$. Then for every $\varepsilon>0$ we can find an $N_b$ such that $|b_n−y|<\varepsilon$ for every $n\geq N_b$. Now let $N=\max\{N_a,N_b\}$.

I can make both terms on the right $\varepsilon/2$.

This is an excellent idea! Yes, in the paragraph above you should replace each $\varepsilon$ with $\varepsilon/2$. Since the definition is true for any number greater than zero, this number can certainly be $\varepsilon/2$.
If you apply this corrections, then you should have a valid proof of the desired claim. To make it more clear, you can start with stating your claim, like this:
Assume $\lim\limits_{n\to\infty}a_n=x$ and $\lim\limits_{n\to\infty}b_n=y$. Let's show that $\lim\limits_{n\to\infty}(a_n+b_n)=x+y$.
